A sample of argon at 300. °C and 50.0 atm pressure is cooled in the same container to a temperature of 0. °C. What is the new pressure?

Question

A sample of argon at 300. °C and 50.0 atm pressure is cooled in the same container to a temperature of 0. °C. What is the new pressure?

105 atm
45.5 atm
54.9 atm
23.8 atm
42.7 atm

Answer 1

First use school boy logic to dismiss any silly answers. If the temperature has dropped then the new pressure must be less than 50.0 atm. Thus we can dismiss 105 atm, and 54.9 atm.

Use P1/T1=P2/T2

or by inspection the temperature has dropped by approximately a factor of 2 (from 573 K to 273 K) so the pressure is approximately half the starting pressure of 50 atm. Thus the answer must 23.8 atm.
[and no calculator needed!]

Answer 2

Well, that sample of argon must be feeling pretty cool. But let's not leave it out in the cold and answer your question!

To solve this problem, we can use the combined gas law equation: P1 * V1 / T1 = P2 * V2 / T2.

The initial pressure (P1) is 50.0 atm at 300 °C (T1). The final temperature (T2) is 0 °C. We're looking for the new pressure (P2).

Plugging in the values, we get: 50.0 atm * V1 / 300 °C = P2 * V2 / 0 °C.

Now, I hate to burst your bubble, but we're missing some important information here. We need to know the initial volume (V1) and the final volume (V2). Without those, we can't calculate the new pressure. So, it looks like the answer is clownishly elusive for now!

Answer 3

To solve this problem, we can use the ideal gas law equation, which states that the pressure (P), volume (V), and temperature (T) of a gas are related by the equation PV = nRT, where n is the number of moles of gas and R is the ideal gas constant.

First, let's find the initial number of moles of argon. We can use the equation PV = nRT, where P is the initial pressure (50.0 atm), V is the volume (which is constant since the gas is in the same container), n is the number of moles of argon, R is the ideal gas constant, and T is the initial temperature (300 °C).

To use the ideal gas law equation, we need to convert the initial temperature from Celsius to Kelvin. Kelvin temperature can be obtained by adding 273.15 to Celsius temperature. So, 300 °C + 273.15 = 573.15 K.

Next, we can rearrange the ideal gas law equation to solve for n:
n = PV / RT

We can plug in the values we know to find the initial number of moles of argon:
n = (50.0 atm)(V) / (0.0821 L.atm/mol.K)(573.15 K)

Now, let's find the final pressure of the argon when cooled to 0 °C. Again, we can use the ideal gas law equation, but this time, the temperature (T) is the final temperature (0 °C) and we want to solve for the final pressure (P).

Like before, we need to convert the final temperature from Celsius to Kelvin. 0 °C + 273.15 = 273.15 K.

Now we can rearrange the ideal gas law equation to solve for the final pressure:
P = nRT / V

We already know n, R, and V, so we can plug in the values we know to find the final pressure of argon:
P = (n)(0.0821 L.atm/mol.K)(273.15 K) / V

Since the volume is constant, we can cancel it out from the equation:
P = n(0.0821 L.atm/mol.K)(273.15 K)

By substituting the value for n that we found earlier, we can calculate the final pressure:
P = [(50.0 atm)(V) / (0.0821 L.atm/mol.K)(573.15 K)](0.0821 L.atm/mol.K)(273.15 K)

Simplifying this expression gives us the final pressure of argon in the same container:
P = 23.8 atm

Therefore, the new pressure of the argon after cooling is 23.8 atm.